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Ok so for example the matrix in REF

\begin{bmatrix} 1 & -2 & 5 & 0 & 3 \\ 0 & 1 & 3 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}

has the row space of \begin{array}{ccccccccccccccc} \{[1 & -2 & 5 & 0 & 3],[0 & 1 & 3 & 0 & 0].[0 & 0 & 0 & 1 & 0]\} \end{array}

which is said to have a dimension of 3. I understand that this is based on the number of matrices, but isn't a "dimension" of 3 supposed to be a 3x1 matrix?

I'm a bit confused as to why they refer to it as the dimmension of the row space.

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The dimension of a vector space $V$ is the number of elements in any basis of $V$.

The row space of an $m\times n$ matrix $A$ is defined as $\operatorname{Row}(A)=\operatorname{Span}\{v_1,\dotsc,v_m\}$ where $v_1,\dotsc,v_m$ are the rows of $A$. Note that $\operatorname{Row}(A)$ is a subspace of $\Bbb R^n$.

Now, if $W$ is a subspace of $\Bbb R^n$ and $\beta=\{w_1,\dotsc,w_k\}$ is a basis for $W$, then $\dim W=k$.

In your case, you have a $4\times 5$ matrix $A$. Note that $\operatorname{Row}(A)$ is a subspace of $\Bbb R^5$ and $$ \beta=\{ \begin{bmatrix} 1&-2&5&0&3 \end{bmatrix}, \begin{bmatrix} 0&1&3&0&0 \end{bmatrix}, \begin{bmatrix} 0&0&0&1&0 \end{bmatrix} \} $$ is a basis for $\operatorname{Row}(A)$. Hence $\dim\operatorname{Row}(A)=3$.

In short, the dimension is a number, not a matrix.

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  • $\begingroup$ Ok, so Row(A) is a subspace of $\Bbb R^5$ and in order for a set to be a basis of a subspace it must span the subspace. How does three 1x5 vecotors span $\Bbb R^5$? For example two 3x1 vectors could not span $\Bbb R^3$. $\endgroup$ – boidkan Apr 17 '14 at 3:52
  • $\begingroup$ @boidkan There is no reason to expect the row-space to span all of $\Bbb R^5$. $\endgroup$ – Brian Fitzpatrick Apr 17 '14 at 3:58
  • $\begingroup$ So the row space is in $\Bbb R^3$ but the matrices are in $\Bbb R^5$? Is this saying that its a three dimensional space somewhere in $\Bbb R^5$? I think I got it. $\endgroup$ – boidkan Apr 17 '14 at 4:00
  • $\begingroup$ @boidkan Not quite. The row space is a three dimensional subspace of $\Bbb R^5$. There is no $\Bbb R^3$ here. $\endgroup$ – Brian Fitzpatrick Apr 17 '14 at 4:06

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